I wanted to know what is the Feynman rules for current-current vertex like this one:

$$ {\cal{ L}} = G^\prime_F \hspace{2mm} \bar{d} \gamma^\mu (1-\gamma^5) u\hspace{3mm} \bar{s}\gamma_\mu (1-\gamma^5)c $$

And what is the simplest way to derive it ? I was trying the following way:
For example in QED, we have $$ {\cal{L}}_{int} = e \bar{\psi}\gamma^\mu \psi A_\mu $$ and omitting the fields (taking partial derivatives) I achieve the vertex:
$$ V_{qed} = i e \gamma^\mu $$ For the current-current case, I wanted to apply the same method, the the Feynman vertex I achieve is this:

$$ V_{current-current} = i G^\prime _F \gamma^\mu (1-\gamma^5) \gamma_\mu (1-\gamma^5) $$ But this appear to be meaningless, because if I simplify this vertex using the anticommutation $ \{\gamma^\mu, \gamma^5 \} = 0 $ and the properties $ (\gamma^5)^2 = 1_{4\times 4} $ and $\gamma^\mu \gamma_\mu = 4 1_{4\times 4} $, then we can see that:

$$ V_{current-current} = i G^\prime _F \gamma^\mu (1-\gamma^5) \gamma_\mu (1-\gamma^5) $$ $$ = i G^\prime _F (1+\gamma^5) \gamma^\mu \gamma_\mu (1-\gamma^5) $$ $$ = 4 i G^\prime _F (1+\gamma^5) (1-\gamma^5) $$ $$ = 4 i G^\prime _F (1 - (\gamma^5)^2) $$ $$ = 4 i G^\prime _F (1 - 1) = 0 $$

What's going on here ? Can anyone please help me find the correct vertex rule !


1 Answer 1


Indeed you find the Feynman rule by differentiating the term in the Lagrangean w.r.t. the fields attached to the vertex.

In the current-current case, you need to be more careful with spinor indices than in the general gauge coupling case. There are four independent spinor indices on such a vertex. You get something proportional to $$ V \propto (\gamma^\mu (1 - \gamma^5))_{ab} (\gamma_\mu (1-\gamma^5))_{cd} $$ where $a, b, c, d$ are spinor indices. Since there is no matrix multiplication between the two parantheses, your Dirac algebra argument fails.

Always, always, always be aware of what indices are suppressed and how these suppressed indices are contracted. In Supersymmetry and GR this will get even more important!

EDIT: The color indices are supressed as well. From group theory we know that $$ \bar{\mathbf{3}} \otimes \mathbf{3} = \mathbf{8} \oplus \mathbf{1}. $$ Also, $$ \mathbf{8} \otimes \mathbf{8} \supset \mathbf{1},$$ so there are two possible color structures here. Assuming that your task is within the standard model and that your coupling strength is the fermi constant I assume that the mediator of your interaction is color neutral (which does not have to be the case in full generality!).

Then, the color indices are contracted within each spinor bilinear and you get the appropriate $\delta_{ab}$. This you need to work out for yourself. Find the contracted indices and the matrix that contracts them, it's super easy for the color indices!

  • $\begingroup$ Thanks. But then what about the color indices ? $\endgroup$
    – quanta
    Aug 12, 2015 at 9:40
  • $\begingroup$ @quanta edited my answer $\endgroup$
    – Neuneck
    Aug 12, 2015 at 9:46

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